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Inessa05 [86]
2 years ago
5

There are slips of paper that are numbered from 1 to 15 in a hat. What is the theoretical probability that a slip of paper that

is chosen will be an even number?
1/15
7/15
1/2
8/15
Mathematics
2 answers:
blsea [12.9K]2 years ago
8 0
Okay! Let's write out all the numbers: 
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

Now, let's bold (or you can circle them... or do really anything to them as long as they are standing out to you) the even numbers (remember all even numbers are divisible by 2)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

Now let's list all the ones we highlighted: 
2
4
6
8
10
12
14

Count up how many numbers that is. 
You should get 7. 

So, our answer is 7 out of 15 numbers. 
Or as a fraction: 

7/15 is our answer.

Note: A probability is the number of chances something has to happen out of the total number of chances. So in this case the number of even numbers out of the total number of numbers. You can apply this to any probability problem. Hope this helps!

DochEvi [55]2 years ago
6 0
There are 7 even numbers between 1 and 15 so the answer will be 7/15

Hope this helps :)
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21. Randy owns a computer store. In 1990, he sold 150 monitors. In 2000, he sold 900
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Answer:

y=75x+150

Step-by-step explanation:

Let y represent the number of monitors sold.

Given:

x is the number of years since 1990.

So, for the year 1990, x=0.

For the year 2000, 10 years passed. So, x=10.

Now, monitors sold in 1990 are 150. So, at x=0,y=150

Monitors sold in 2000 are 900. So, at x=10,y=900

Thus, the two points are (0,150) and (10,900).

The slope of a line with points (x_{1},y_{1}) and (x_{2},y_{2})  is given as:

Slope, m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

For the points (0,150) and (10,900), the slope is given as:

Slope, m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ m=\frac{900-150}{10-0}\\ m=75

Now, the y-intercept is the point where x=0. So, the point (0,150) has x value as 0.

So, the y-intercept is 150.

Equation of a line in slope-intercept form is given as:

y=mx+b

Where, m is the slope and b is the y intercept.

Here, m=75 and b=150.

Therefore, the equation that represents the above data is given as:

y=75x+150

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2 years ago
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Step-by-step explanation:

In the question some data is missing, that's value can be defined as follows:

Values:

\ X =  3.3 \\

\mu \ = 2.8 \% \\\sigma = 0.6

Formula:

z= \frac{X - \mu }{\sigma} \\

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2 years ago
Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that (a)
UNO [17]

Answer:

a) P=0.226

b) P=0.6

c) P=0.0008

d) P=0.74

Step-by-step explanation:

We know that the seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Therefore, we have 46 balls.

a) We calculate the probability that are 3 red, 2 blue, and 2 green balls.

We calculate the number of possible combinations:

C_7^{46}=\frac{46!}{7!(46-7)!}=53524680

We calculate the number of favorable combinations:

C_3^{12}\cdot C_2^{16}\cdot C_2^{18}=660\cdot 120\cdot 153=12117600

Therefore, the probability is

P=\frac{12117600}{53524680}\\\\P=0.226

b) We calculate the probability that are at least 2 red balls.

We calculate the probability  withdrawn of 1 or none of the red balls.

We calculate the number of possible combinations:

C_7^{46}=\frac{46!}{7!(46-7)!}=53524680

We calculate the number of favorable combinations: for 1 red balls

C_1^{12}\cdot C_7^{34}=12\cdot 1344904=16138848

Therefore, the probability is

P_1=\frac{16138848}{53524680}\\\\P_1=0.3

We calculate the number of favorable combinations: for none red balls

C_7^{34}=5379616

Therefore, the probability is

P_0=\frac{5379616}{53524680}\\\\P_0=0.1

Therefore, the  the probability that are at least 2 red balls is

P=1-P_1-P_0\\\\P=1-0.3-0.1\\\\P=0.6

c) We calculate the probability that are all withdrawn balls are the same color.

We calculate the number of possible combinations:

C_7^{46}=\frac{46!}{7!(46-7)!}=53524680

We calculate the number of favorable combinations:

C_7^{12}+C_7^{16}+C_7^{18}=792+11440+31824=44056

Therefore, the probability is

P=\frac{44056}{53524680}\\\\P=0.0008

d) We calculate the probability that are either exactly 3 red balls or exactly 3 blue balls are withdrawn.

Let X, event that exactly 3 red balls selected.

P(X)=\frac{C_3^{12}\cdot C_4^{34}}{53524680}=0.57\\

Let Y, event that exactly 3 blue balls selected.

P(Y)=\frac{C_3^{16}\cdot C_4^{30}}{53524680}=0.29\\

We have

P(X\cap Y)=\frac{18\cdot C_3^{12} C_3^{16}}{53524680}=0.12

Therefore, we get

P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\\\P(X\cup Y)=0.57+0.29-0.12\\\\P(X\cup Y)=0.74

8 0
2 years ago
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